Question

我列出了一些对象 - 让我们假设公司。现在，我想检查此列表是否包含具有某些名称但未考虑订单的公司。目前我正在使用这样的结构：

companyList.name.sort() == ["First", "Second"]

Spock或Groovy中是否有任何运算符允许我比较没有顺序的数组？

Answer 1

据我所知，没有这样的操作员。如果列表不包含任何重复项，则可以使用以下断言：

companyList.name as Set == ["First", "Second"] as Set

或类似的东西：

companyList.name.size() == ["First", "Second"].size() && companyList.name.containsAll(["First", "Second"])

Answer 2

您可以在Spock中使用Hamcrest支持，并使用为此案例明确设计的Matcher - containsInAnyOrder。您需要以下导入：

import static org.hamcrest.Matchers.containsInAnyOrder
import static spock.util.matcher.HamcrestSupport.that

然后您可以按如下方式编写测试代码：

given:
    def companyList = [ "Second", "First"]
expect:
    that companyList, containsInAnyOrder("First", "Second")

这比使用.sort()有好处，因为列表中的重复元素将被正确考虑。以下测试将使用Hamcrest失败，但使用.sort()

传递

given:
    def companyList = [ "Second", "First", "Second"]
expect:
    that companyList, containsInAnyOrder("First", "Second")

Condition not satisfied:

that companyList, containsInAnyOrder("First", "Second")
|    |
|    [Second, First, Second]
false

Expected: iterable over ["First", "Second"] in any order
     but: Not matched: "Second"

如果您使用then:代替expect:，则可以使用expect代替that导入以提高可读性。

then:
    expect companyList, containsInAnyOrder("First", "Second")

Answer 3

为@Opal回答，它可能是最好的。作为好奇心，我只会添加减号运算符的使用示例来比较两个未排序的列表：

import spock.lang.Specification

class ComparingListsSpec extends Specification {

    def "should contain all companies in two unsorted lists"() {
        when:
        def companies = ["Lorem", "ipsum", "dolor", "sit", "amet"]
        def unsorted = ["ipsum", "sit", "Lorem", "dolor", "amet"]

        then:
        companies - unsorted == []

        and:
        unsorted - companies == []

        and:
        companies - unsorted in [[]]
    }
}

如果其中一个列表包含冗余数据，它也可以工作。

Spock中的数组断言

3 个答案: